PHY646 - Quantum Field Theory and the Standard Model

Even Term 2020 Dr. Anosh Joseph, IISER Mohali

LECTURE 17

Monday, February 10, 2020

Topics: Abelian Gauge Invariance, Covariant Derivatives.

Abelian Gauge Invariance

In the absence of any sources, we have the Lagrangian for Maxwell’s equations

1 L = − F F µν, (1) 4 µν where the field strength is defined by

Fµν = ∂µAν − ∂νAµ. (2)

This Lagrangian has an important property known as gauge invariance or gauge symmetry. The Lagrangian is invariant under the transformation

Aµ(x) −→ Aµ(x) + ∂µα(x), (3) for any function α(x), which dies off conveniently fast at spatial infinity (x → ∞). The field strength tensor is invariant under the gauge symmetry

0 0 0 Fµν = ∂µAν − ∂νAµ

= ∂µ(Aν + ∂να) − ∂ν(Aµ + ∂µα)

= ∂µAν + ∂µ∂να − ∂νAµ − ∂ν∂µα

= Fµν. (4)

This is telling us that the Maxwell Lagrangian has an infinite number symmetries corresponding to each function α(x). This symmetry is a local symmetry, instead of a global symmetry, since the function α is spacetime dependent. PHY646 - Quantum Field Theory and the Standard Model Even Term 2020

When a system has a gauge symmetry, the system has a redundancy in it. We can say that two states of the system that are related to each other by a gauge symmetry are identical. That is, they are the same physical state. The equations of motion following from our Lagrangian, the Maxwell’s equations,

ν Aµ − ∂µ(∂νA ) = 0 (5) are not enough to specify the evolution of Aµ. The operator Gµν = gµν − ∂µ∂ν (6) is not invertible. It can be easily seen in momentum space. The object

2 −gµνp + pνpµ, (7) has a zero eigenvalue, with eigenvector pµ and thus

2 det (−p gµν + pµpν) = 0. (8)

What this means is that we are unable to find a unique solution for Aµ. Suppose we are given Aµ as initial data. Then there is no way to uniquely determine Aµ at a later time since we are not able to distinguish between Aµ and Aµ + ∂µα. µ The operator Gµν annihilates any function of the form ∂ α. We have

µ µ µ Gµν(∂ α) = gµν∂ α − ∂µ∂ν∂ α µ = ∂να − ∂µ∂ ∂να = 0. (9)

Thus we are facing a problem. If we are happy with identifying Aµ and Aµ+∂µα as corresponding to the same physical state, then our trouble is over. In electromagnetism, we can think of our phase space as an enlarged one, foliated by gauge orbits. The states that lie along a given curve can be reached by a gauge transformation and are identified. We can select a representative from each gauge orbit. It does not matter which representative we choose from each gauge orbit since they are all equivalent. The manner in which we “slice” the gauge orbits leads to different gauge choices. See Fig. 1. There are many possibilities for gauge choices. We can think of choosing a gauge as picking coordinates that are adapted to a particular problem. The set of equations given in Eq. (5) is really four equations. Let us separate out the 0 and i

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Gauge fixing

Gauge orbits

Figure 1: A schematic diagram showing the gauge fixing procedure. The families of gauge invariant configurations are represented by the gauge orbits. Gauge fixing corresponds to a slice that intersects each orbit once. components. We have

2 −∂j A0 + ∂0∂jAj = 0, (10)

Ai − ∂i(∂0A0 − ∂jAj) = 0. (11)

Let us count the physical degrees of freedom associated with the field Aµ. We will use the freedom of transforming the fields in Eq. (3) to impose constraints on Aµ and thus fix the gauge.

Coulomb Gauge

Let us choose the constraint 5~ · A~ = 0, (12) which is expressed in components as ∂iAi = 0.

In the Maxwell Lagrangian we see that the field A0 has no kinetic term: A˙ 0 is absent in the

Lagrangian. That is the field A0 is not dynamical.

This means that if we are given some initial data Ai and A˙ i at a time t0, then the field A0 is fully determined by the equation of motion

5~ · E~ = 0, (13)

3 / 8 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 which, expanding out reads

∂iFi0 = 0,

∂i(∂iA0 − ∂0Ai) = 0, 2 5 A0 + 5 · ∂0A~ = 0. (14)

This has the solution Z 1 A (~x) = d3x0 5~ · ∂ A~(~x0). (15) 0 4π|~x − ~x0| t

Thus A0 is not independent: we do not get to specify A0 on the initial time slice. Since we have the gauge condition 5~ · A~ = 0 from Eq. (15) we see that we can set

A0 = 0. (16)

Thus we have reduced the number of independent degrees of freedom in Aµ down to 3.

We can also derive the constraint A0 = 0 using the following. The gauge transformations

Aµ → Aµ + ∂µα (17) is respected by the Coulomb gauge ∂iAi = 0 if in the spatial part of the gauge tranaformation

Ai → Ai + ∂iα (18) we have the following constraint on α: 2 ∂i α = 0. (19)

Thus we have

2 ∂iAi → ∂iAi + ∂i α

∂iAi → ∂iAi + 0 (20)

2 That is the constraint ∂i α = 0 is needed for the Coulomb gauge ∂iAi = 0 to respect the spatial part of the gauge transformations. We are not done yet. How about the temporal part of the gauge transformations? Let us come back to the Maxwell’s equations, with the 0 and i components separated out

2 −∂j A0 + ∂t∂jAj = 0,

Ai − ∂i(∂tA0 − ∂jAj) = 0.

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Applying Coulomb gauge ∂iAi = 0, we have

2 ∂j A0 = 0,

Ai − ∂i∂0A0 = 0.

It is always possible to find a gauge transformation α

0 A0 = A0 + ∂0α (21)

0 such that A0 = 0. The choice is ∂0α = −A0, (22) which is consistent provided A0 vanishes for x0 → ±∞.

Thus we have eliminated one degree of freedom from Aµ completely, and we are down to three. One more to go.

In Coulomb gauge (∂iAi = 0 and A0 = 0) Maxwell’s equations reduce to

Ai = 0. (23)

These equations seem to propagate three modes. But let us not forget that Ai is constrained by

∂iAi = 0. (24)

Taking the Fourier transform we have

Z d4p A (x) =  (p) eipx. (25) µ (2π)4 µ

The equations become

p2 = 0 (equations of motion), (26)

pii = 0 (gauge choice), (27)

0 = 0 (gauge choice). (28)

Let us choose a frame in which we can write the momentum as

pµ = (E, 0, 0,E). (29)

Then the above set of equations give two solutions

(1) µ = (0, 1, 0, 0), (30) (2) µ = (0, 0, 1, 0), (31)

5 / 8 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 which represent linearly polarized light. Thus we have constructed a theory propagating only two degrees of freedom, representing a massless spin-1 particle. Another common basis for the transverse polarizations of light is

1 R = √ (0, 1, i, 0), (32) µ 2 1 L = √ (0, 1, −i, 0). (33) µ 2

These polarizations correspond to circularly polarized light and are called helicity eigenstates.

Lorenz Gauge

We could also use the gauge choice

∂µAµ = 0, (34) called the Lorenz gauge. In this gauge we can find three vectors that satisfy

µ p µ = 0. (35)

They are

(1) µ = (0, 1, 0, 0), (36) (2) µ = (0, 0, 1, 0), (37) (f) µ = (1, 0, 0, 1). (38)

The first two modes are the physical transverse polarizations. The third apparent solution (f) denoted by µ is called the forward polarization. It does not correspond to a physical state. (f) One way to see this is to note that µ is not normalizable

∗  (f) (f) µ µ = 0. (39)

Another way to note is that (f) µ ∝ pµ, (40) which corresponds to

Aµ = ∂µφ, (41) for some φ. This field configuration is gauge-equivalent to

Aµ = 0. (42)

That is, choosing α = −φ in the gauge transformation equation. Thus the forward polarization

6 / 8 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 corresponds to a field configuration that is pure gauge.

Covariant Derivatives

Let us now think about adding interactions to the gauge invariant Lagrangian, keeping in mind that any term we add to the Lagrangian must respect gauge invariance. An interaction term like

Lint = Aµφ∂µφ, (43) is not gauge invariant. Under the gauge transformation it picks up extra term

Aµφ∂µφ → Aµφ∂µφ + (∂µα)φ∂µφ. (44)

It is impossible to couple Aµ to any field with only one degree of freedom, such as the real scalar field φ given above. We must be able to make the field φ transform to compensate for the gauge transformation of Aµ, in order to cancel the ∂µα term. But if there is only one field φ, it has nothing to mix with so it cannot transform.

Thus, we need at least two fields φ1 and φ2. It is easiest to deal with such a doublet by putting them together into a complex field

φ = φ1 + φ2, (45) and then to work with φ and φ∗. Under gauge transformation, φ can transform as

φ → e−iα(x)φ, (46) which makes m2φ∗φ (47) gauge invariant. But a derivative term like

2 |∂µφ| (48) is not invariant. We can make this kinetic term gauge invariant using something we call a covariant derivative.

Adding a conventional constant e to the transformations of Aµ, so

1 A → A + ∂ α, (49) µ µ e µ

7 / 8 PHY646 - Quantum Field Theory and the Standard Model Even Term 2020 we find

−iα(x) (∂µ + ieAµ)φ → (∂µ + ieAµ + i∂µα)e φ −iα(x) = e (∂µ + ieAµ)φ. (50)

In the line above we have moved the e−iα(x) factor to the left through the partial derivative. This leads us to define the covariant derivative as

Dµφ ≡ (∂µ + ieAµ)φ. (51)

It transforms just like the field φ does

−iα(x) Dµφ → e Dµφ. (52)

Thus the Lagrangian

1 L = − F F µν + (D φ)∗(D φ) − m2φ∗φ (53) 4 µν µ µ is gauge invariant. This is the Lagrangian for scalar QED.

In general, different fields φn can have different charges Qn and they transform as

Qniα(x) φn → e φn. (54)

Then the covariant derivative is

Dµφn = (∂µ − ieQnAµ)φn. (55)

Note that earlier we took Q = −1 for φ thinking of it as an electron with charge −1. Symmetries parametrized by a function such as α(x) are called gauge or local symmetries, while if they are only symmetries for constant α they are called global symmetries. For gauge symmetries, we can pick a separate transformation at each point in spacetime. A gauge symmetry automatically implies a global symmetry.

References

[1] M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press (2013).

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